A) \[\frac{{{\pi }^{2}}}{6}\]
B) \[\frac{{{\pi }^{2}}}{12}\]
C) \[\frac{{{\pi }^{2}}}{24}\]
D) None of these
Correct Answer: B
Solution :
| [b] Let \[f(x)=\frac{\left| x \right|}{8{{\cos }^{2}}2x+1}\] |
| then \[f(-x)=\frac{\left| -x \right|}{8{{\cos }^{2}}2(-x)+1}=\frac{\left| x \right|}{8{{\cos }^{2}}2x+1}\] |
| \[=f(x)\] |
| \[\therefore f(x)\] is even function |
| \[\therefore I=\int\limits_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\frac{\left| x \right|dx}{8{{\cos }^{2}}2x+1}=2\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left| x \right|dx}{8{{\cos }^{2}}2x+1}}}\] |
| \[=2\int\limits_{0}^{\frac{\pi }{2}}{\frac{xdx}{8{{\cos }^{2}}2x+1}=2{{I}_{1}}}\] |
| Now |
| \[\therefore {{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left( \frac{\pi }{2}-x \right)dx}{8{{\cos }^{2}}2\left( \frac{\pi }{2}-x \right)+1}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\frac{\pi }{2}-x}{8{{\cos }^{2}}2x+1}dx}}\] |
| \[=\frac{\pi }{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{dx}{8{{\cos }^{2}}2x+1}-{{I}_{1}}}\] |
| \[2{{I}_{1}}=\frac{\pi }{2}\,\,.\,\,2\int\limits_{0}^{\frac{\pi }{4}}{\frac{dx}{8{{\cos }^{2}}2x+1}}\] |
| \[=\pi \int\limits_{0}^{\frac{\pi }{4}}{\frac{{{\sec }^{2}}2x}{9+{{\tan }^{2}}2x}dx;}\] |
| Put \[\tan 2x=t\Rightarrow 2{{\sec }^{2}}2xdx=dt\] |
| \[2{{I}_{1}}=\frac{\pi }{2}\int\limits_{0}^{\infty }{\frac{dt}{9+{{t}^{2}}}=\frac{\pi }{2}.\frac{1}{3}\left[ {{\tan }^{-1}}\frac{t}{3} \right]_{0}^{\infty }=\frac{\pi }{2}.\frac{1}{3}.\frac{\pi }{2}}\] |
| \[\therefore {{I}_{1}}=\frac{{{\pi }^{2}}}{24}\Rightarrow I=2{{I}_{1}}=\frac{{{\pi }^{2}}}{12}\] |
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